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Physics - Light (Reflection, Spherical Mirrors)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Reflection of Light: The phenomenon of bouncing back of light into the same medium when it strikes a polished surface. The Laws of Reflection state that i=r\angle i = \angle r and the incident ray, reflected ray, and the normal at the point of incidence all lie in the same plane.

Spherical Mirrors: These are mirrors whose reflecting surfaces are parts of a hollow sphere. A Concave mirror is silvered on the outer surface (converging), while a Convex mirror is silvered on the inner surface (diverging).

Principal Terms: The geometric center of the mirror is the Pole (PP). The center of the hollow sphere is the Center of Curvature (CC). The distance PCPC is the Radius of Curvature (RR).

Principal Focus (FF): For a concave mirror, it is a point on the principal axis where rays parallel to the axis actually meet after reflection. For a convex mirror, rays appear to diverge from this point.

Focal Length (ff): The distance between the pole PP and the principal focus FF. For spherical mirrors of small aperture, the focal length is half the radius of curvature: f=R2f = \frac{R}{2}.

New Cartesian Sign Convention: All distances are measured from the Pole (PP). Distances in the direction of incident light are positive; opposite to incident light are negative. Heights above the principal axis are positive (+h+h), and below are negative (h-h).

Magnification (mm): The ratio of the height of the image (hih_i) to the height of the object (hoh_o). If m>1|m| > 1, the image is magnified; if m<1|m| < 1, it is diminished. A negative mm indicates a real and inverted image, while a positive mm indicates a virtual and erect image.

📐Formulae

f=R2f = \frac{R}{2}

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

m=hiho=vum = \frac{h_i}{h_o} = -\frac{v}{u}

💡Examples

Problem 1:

An object is placed at a distance of 30 cm30\text{ cm} from a concave mirror of focal length 20 cm20\text{ cm}. Find the position and nature of the image formed.

Solution:

Given: u=30 cmu = -30\text{ cm}, f=20 cmf = -20\text{ cm}. Using the mirror formula: 1v+1u=1f1v=1f1u\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}. Substituting values: 1v=120130=3+260=160\frac{1}{v} = \frac{1}{-20} - \frac{1}{-30} = \frac{-3 + 2}{60} = -\frac{1}{60}. Therefore, v=60 cmv = -60\text{ cm}. Magnification m=vu=6030=2m = -\frac{v}{u} = -\frac{-60}{-30} = -2.

Explanation:

Since vv is negative, the image is formed 60 cm60\text{ cm} in front of the mirror. Since mm is negative and greater than 11, the image is real, inverted, and magnified (twice the size of the object).

Problem 2:

A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m3.00\text{ m}. If a bus is located at 5.00 m5.00\text{ m} from this mirror, find the position of the image.

Solution:

Given: R=+3.00 mR = +3.00\text{ m}, u=5.00 mu = -5.00\text{ m}. Focal length f=R2=3.002=+1.50 mf = \frac{R}{2} = \frac{3.00}{2} = +1.50\text{ m}. Using 1v=1f1u=11.5015.00=11.5+15=5+1.57.5=6.57.5\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1.50} - \frac{1}{-5.00} = \frac{1}{1.5} + \frac{1}{5} = \frac{5 + 1.5}{7.5} = \frac{6.5}{7.5}. Thus, v=7.56.5=+1.15 mv = \frac{7.5}{6.5} = +1.15\text{ m}.

Explanation:

The image is formed at a distance of 1.15 m1.15\text{ m} behind the mirror. The positive sign indicates that the image is virtual and erect.

Light (Reflection, Spherical Mirrors) Revision - Class 9 Science ICSE